### Cubic Spin

Prove that the graph of f(x) = x^3 - 6x^2 +9x +1 has rotational symmetry. Do graphs of all cubics have rotational symmetry?

### Sine Problem

In this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern.

### Parabolic Patterns

The illustration shows the graphs of fifteen functions. Two of them have equations y=x^2 and y=-(x-4)^2. Find the equations of all the other graphs.

# Ellipses

##### Stage: 4 and 5 Challenge Level:

Well done Ryan and Prateek from Riccarton High School, Christchurch, New Zealand and Andrei from School 205, Bucharest, Romania for your solutions.

Ryan noticed that the formula x 2 / 36 + y 2 / 16 = 1 gives a large ellipse crossing through points 6 and -6 on the x axis and points 4 and -4 on the y axis. From this he observed that the formula contains (x 2 / 36) and 36 is a square of 6 and -6. When x takes these values then y=0. This also works for the part of the formula (y 2 / 16) as 16 is a square of 4 and -4. When y takes these values then x=0.

Andrei explained how he found the other equations as follows.

I represented first the curve: $$x^2 + y^2 = 1 \quad (1)$$

For this, I observed that both x and y could have values between -1 and +1. I consider x as the independent variable, and from the eq. (1) I determined y: $$y = \pm \sqrt{1 - x^2}$$

I gave values to x, from -1 to 1, step 0.1, and I calculated y. I had to calculate two sets of values for y, one corresponding to the plus sign, the other to the minus sign. Then I plotted y as a function of x for both, and I obtained the circle in the middle.

For the curve $$\frac {x^2}{36} + \frac {y^2}{16} = 1 \quad (2)$$

I considered again x as the independent variable. It varies between -6 to 6. The equation for y is: $$y = \pm \sqrt{16( 1 - \frac {x^2}{36})}= \pm4 \sqrt {1 - ( \frac{x^2}{36})}$$

It is visible even from the equation that y varies between -4 and 4.

Now, as I understood that in the general equation: $$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \quad (3)$$

x varies between -a and a, and y between -b and b, I drew all other curves in the same manner. The equations of the other 8 graphs are: $$\frac {x^2}{7^2} + \frac {y^2}{4^2} = 1 \quad (4)$$ $$\frac {x^2}{5^2} + \frac {y^2}{4^2} = 1 \quad (5)$$ $$\frac {x^2}{4^2} + \frac {y^2}{4^2} = 1 \quad (6)$$ or , i.e. a circle of radius 4. $$\frac {x^2}{3^2} + \frac {y^2}{4^2} = 1 \quad (7)$$ $$\frac {x^2}{2^2} + \frac {y^2}{4^2} = 1 \quad (8)$$ $$\frac {x^2}{1^2} + \frac {y^2}{4^2} = 1 \quad (9)$$ $$\frac {x^2}{6^2} + \frac {y^2}{3^2} = 1 \quad (10)$$ $$\frac {x^2}{7^2} + \frac {y^2}{1^2} = 1 \quad (11)$$

All these ellipses are symmetrical about both x and y axes because by changing x to -x and/or y to -y the equation (3) doesn't change.